Indija mo 2011

Put $x=y=0$; we get $f(0{)}^2=4f(0{)}^2$ and hence $f(0)=0$.

$$f(y)(f(y)-f(-y))=0.$$

We may conclude that either $f(y)=0$ or $f(y)=f(-y)$ for each $y\in \mathbb{R}$. Replacing $y$ by $-y$, we may also conclude that $f(-y)(f(-y)-f(y))=0$. If $f(y)=0$ and $f(-y)\ne 0$ for some $y$, then we must have $f(-y)=f(y)=0$, a contradiction. Hence either $f(y)=f(-y)=0$ or $f(y)=f(-y)$ for each $y$. This forces $f$ is an even function.

Taking $y=1$ in (1), we get $$f(x+1)f(x-1)=(f(x)+f(1){)}^2-4x^{2}f(1).$$

Replacing $y$ by $x$ and $x$ by $1$, you also get $$f(1+x)f(1-x)=(f(1)+f(x){)}^2-4f(x).$$

Comparing these two using the even nature of $f$, we get $f(x)=c{x}^2$, where $c=f(1)$. Putting $x=y=1$ in (1), you get $4c^2-4c=0$. Hence $c=0$ or $1$. We get $f(x)=0$ for all $x$ or $f(x)=x^2$ for all $x$.